3个回答
NHK : 因为 (1 r r^2 r^3 ... r^n)*(1 - r)
= (1 r r^2 r^3 ... r^n) - [r r^2 r^3 r^4 ... r^(n 1)]
由于 r, r^2, r^3...等等....都可以互相減去.......
(1 r r^2 r^3 ... r^n)*(1 - r) = [1 - r^(n 1)]
所以 1 r r^2 r^3 ... r^n = [1 - r^(n 1)] / (1 - r)
或改写成 r^0 r^1 r^2 r^3 ... r^n = [1 - r^(n 1)] / (1 - r)
原式 = lim n->无限大 1/2^0 1/2^1 1/2^2 ... 1/2^n / 1/3^0 1/3^1 1/3^2 ... 1/3^n
1 1/2 1/4 .... 1/2^n is a geometic series and the common difference is 1/2.
Similarly, 1 1/3 1/9 .... 1/3^n is a geometic series and the common difference is 1/3.
原式化简后 = lim n->无限大 {[1 - 1/2^(n 1)] / (1 - 1/2)} / {[1 - 1/3^(n 1)] / (1 - 1/3)}
= lim n->无限大 [1 - 1/2^(n 1)]*(1 - 1/3) / (1 - 1/2)*[1 - 1/3^(n 1)]
= lim n->无限大 [1 - 1/2^(n 1)]*(2/3) / (1/2)*[1 - 1/3^(n 1)]
= lim n->无限大 4/3*[1 - 1/2^(n 1)] / [1 - 1/3^(n 1)]
由于 lim n->无限大 1/2^(n 1) = 0 和 lim n->无限大 1/3^(n 1) = 0
= lim n->无限大 4/3*(1 - 0) / (1 - 0)
= 4/3 <----- 最终答案
运用罗比达法则后，利用等价无穷小即可。
谢谢，一看到x=1/t就想起来了，怎么当初就是没想到呢！！哎，还是做题太少啊！！thank you very much!